Answer to Question #107174 in Statistics and Probability for jasmine

Question

Health Canada wishes to estimate the mean amount of mercury μ, in parts per million (ppm), in commercial fish. Specifically, they want to estimate the value of μ to within 0.03, and be 98% confident in their results. Assume that the amount of mercury in commercial fish varies from one fish to another, the standard deviation being σ=0.4 ppm. How many commercial fish should be randomly chosen, each chosen will have the mercury content measured, to satisfy Health Canadas estimation of μ?

Answer 1

Provided that a desired margin of error $E$ is given, the population standard deviation $\sigma$ is provided, and the significance level is specified, we can compute the minimum required sample size that will lead to a margin of error less than or equal to the one specified, by using the following formula:

$$n\geq({z_c\sigma \over E})^2$$

The critical value for $\alpha=0.02$ is $z_c=z_{1-\alpha/2}=2.326$

Given that $E=0.03, \sigma=0.4.$ Then

$$n\geq({2.326\cdot0.4 \over 0.03})^2$$

$$n\geq962$$